Question: Solve for $x$ : $5x^2 - 90x + 405 = 0$
Answer: Dividing both sides by $5$ gives: $ x^2 {-18}x + {81} = 0 $ The coefficient on the $x$ term is $-18$ and the constant term is $81$ , so we need to find two numbers that add up to $-18$ and multiply to $81$ The number $-9$ used twice satisfies both conditions: $ {-9} + {-9} = {-18} $ $ {-9} \times {-9} = {81} $ So $(x - {9})^2 = 0$ $x - 9 = 0$ Thus, $x = 9$ is the solution.